The base of a solid $S$ is the region bounded by the $x$ -axis, the $y$ -axis, and the line through the points $(0,a)$ and $(b,0)$. $y$ $x$ $(0,a)$ $(b,0)$ Cross-sections perpendicular to the $y$ -axis are equilateral triangles. Determine the exact volume of solid $S$.
Explanation: Let's graph the base of the solid, shaded in blue. The thin orange rectangle depicts a representative cross-section sitting on the base. The length of the green segment is $x$. $y$ $x$ $(0,a)$ $(b,0)$ $(x,y)$ $x$ Since each cross-section is perpendicular to the $y$ -axis, the independent variable is $y$. We can see from the graph that $y$ goes from $0$ to $a$. If $A$ denotes the area of each cross-section as a function of $y$, the volume $V$ of solid $S$ is $ V=\int_0^a A(y) \,dy$. To determine the area $A$ as a function of $y$, first express $A$ in terms of $x$. Since the triangular cross-section rests on the rectangle pictured above, the length of the base of the triangle is $x$. Since the triangle is equilateral, the height is $\sqrt3x/2$. $x$ $\dfrac{\sqrt3}2x$ The area $A$ of the triangle is $A=\dfrac12\cdot x\cdot\dfrac{\sqrt3}2x=\dfrac{\sqrt3}4x^2$. What is $A$ as a function of $y$ ? The corner point $(x,y)$ of the rectangle lies on the line through the points $(0,a)$ and $(b,0)$. The equation of this line is $\dfrac xb+\dfrac ya=1$. Let's solve the equation for $x$ : $x=b\left(1-\dfrac ya\right)$. Now we can express $A=\dfrac{\sqrt3}4x^2$ in terms of $y$ as $\begin{aligned} A(y)&=\dfrac{\sqrt3}4b^2\left(1-\dfrac ya\right)^2 \\\\ &=\dfrac{\sqrt3}4b^2\left(\dfrac ya-1\right)^2 \end{aligned}$ Can you express the volume $V$ of solid $S$ as a definite integral? The volume formula gives us the definite integral $\begin{aligned} V&=\int_0^a A(y) \,dy \\\\ &=\int_0^a \dfrac{\sqrt3}4b^2\left(\dfrac ya-1\right)^2 dy \\\\ &=\dfrac{\sqrt3}4b^2\int_0^a \left(\dfrac ya-1\right)^2 dy \end{aligned}$ What is the value of the integral? $\begin{aligned} V&=\dfrac{\sqrt3}4b^2\int_0^a \left(\dfrac ya-1\right)^2 dy \\\\ &=\dfrac{\sqrt3}4b^2\left[\dfrac a3\left(\dfrac ya-1\right)^3\right]_0^a \\\\ &=\dfrac{\sqrt3}4b^2\cdot\dfrac a3\left[\left(\dfrac aa-1\right)^3-\left(\dfrac 0a-1\right)^3\right] \\\\ &=\dfrac{ab^2}{4\sqrt3}\left[(0)^3-(-1)^3\right] \\\\ &=\dfrac{ab^2}{4\sqrt3} \end{aligned}$